Question

lock of mass m, 3.08 kg on a frictionless plane incined at angle 0 29.1. is connected by a cord over a massless, frictionless puley to a second block of mass m2-2.34 kg hanging vertic (see the figure). (a) What is the acceleration of the hanging block (choose the positive direction down)? (b) What is the tension in the cord? (a) Number Units (b) Number Units

Answer 1

Using Newton's second law, we have Sum of the forces on the block system = mass of the system times the acceleration of the system:
- (m2) g + (m1) g cos(90 - 29.1) = (m1 + m2) a
Note: - (m2) g is the weight force on the hanging block pointing down;
(m1) g cos(90 - 29.1) is the component of m1's weight force pointing down the incline;
m1 + m2 is the mass of the system
a = acceleration of the system

- (m2) g + (m1) g cos(90 - 29.1) = (m1 + m2) a
- (2.34 kg)(9.81 m/s^2) + (3.08 kg)(9.81 m/s^2) cos(60.9 degrees) = (2.34 kg + 3.08 kg) a
a = -1.345 m/s^2
Note: since a is positive the hanging block moves downward

(b.) The tension in the cord can be obtained by looking at Newton's Second Law for either block. Looking at the hanging block we have:
- (m2) g + T = (m2) a where T = tension in cord pointing up
- (2.34 kg)(9.81 m/s^2) + T = (2.34 kg)(1.345 m/s^2)
T = 26.102 Newtons